9 Replacement problems.pptx

1. (c) 2001 Contemporary Engineering Economics 1
Chapter 15
Replacement Decisions
• Replacement Analysis
Fundamentals
• Economic Service Life
• Replacement Analysis When
Required Service is Long
• Replacement Analysis with Tax
Consideration

2. (c) 2001 Contemporary Engineering Economics 2
Replacement Terminology
• Sunk cost: any past cost
unaffected by any future
decisions
• Trade-in allowance: value
offered by the vendor to reduce
the price of a new equipment
• Operating Cost
• Defender: an old machine
• Challenger: new machine
• Current market value: selling
price of the defender in the
market place

3. (c) 2001 Contemporary Engineering Economics 3
Sunk Cost associated with an Asset’s Disposal
$0 $5000 $10,000 $15,000 $20,000 $25,000 $30,000
Original investment
$10,000 $5000
Market value
$10,000
Lost investment
(economic depreciation) Repair cost
$20,000
Sunk costs = $15,000

4. (c) 2001 Contemporary Engineering Economics 4
Replacement Decisions
• Cash Flow Approach
• Treat the proceeds from sale of the old
machine as down payment toward
purchasing the new machine.
• Can be used if the analysis period is
same for all alternatives.
• Use NPW or AE analysis to decide
• Opportunity Cost Approach
• Treat the proceeds from sale of the old
machine as the investment required to
keep the old machine.

5. (c) 2001 Contemporary Engineering Economics 5
Replacement Analysis – Cash Flow
Approach
0 1 2 3
0 1 2 3
$8000
$2500
$15,000
$6000
$5500
(a) Defender (b) Challenger
$10,000
Sales proceeds
from defender

6. (c) 2001 Contemporary Engineering Economics 6
Annual Equivalent Cost - Cash Flow
Approach
 Defender:
PW(12%)D = $2,500 (P/F, 12%, 3) - $8,000 (P/A, 12%, 3)
= - $17,434.90
AE(12%)D = PW(12%)D(A/P, 12%, 3)
= -$7,259.10
 Challenger:
PW(12%)C = $5,500 (P/F, 12%, 3) - $5,000
- $6,000 (P/A, 12%, 3)
= -$15,495.90
AE(12%)C = PW(12%)C(A/P, 12%, 3)
= -$6,451.79
Replace the
defender
now!

7. (c) 2001 Contemporary Engineering Economics 7
Opportunity Cost Approach
0 1 2 3 0 1 2 3
$8000
$2500
$15,000
$6000
$5500
Defender Challenger
$10,000
Proceeds from sale viewed as
an opportunity cost of keeping
the asset

8. (c) 2001 Contemporary Engineering Economics 8
 Defender:
PW(12%)D = -$10,000 - $8,000(P/A, 12%, 3) + $2,500(P/F, 12%, 3)
= -$27,434.90
AE(12%)D = PW(12%)D(A/P, 12%, 3)
= -$11,422.64
 Challenger:
PW(12%)C = -$15,000 - $6,000(P/A, 12%, 3) + $5,500(P/F, 12%, 3)
= -$25,495.90
AE(12%)C = PW(12%)C(A/P, 12%, 3)
= -$10,615.33
Opportunity Cost Approach
Replace the
defender now!

12. (c) 2001 Contemporary Engineering Economics 12
Economic Service Life
• Def:Economic service life is the
useful life of a defender, or a
challenger, that results in the
minimum equivalent annual cost
• Why do we need it?: We should use
the respective economic service lives
of the defender and the challenger
when conducting a replacement
analysis.
Ownership (Capital)
Cost (init.+salvg.)
Operating
cost
+
Minimize
CR i I A P i N S A F i N
N
( ) ( / , , ) ( / , , )
 

13. (c) 2001 Contemporary Engineering Economics 13
Mathematical Relationship
• Capital Recov. Cost.
• Operating Cost:
• Total Cost:
• Objective: Find n* that
minimizes AEC
CR i I A P i N S A F i N
N
( ) ( / , , ) ( / , , )
 
OC i OC P F i n A P i N
n
n
N
( ) ( / , , ) ( / , , )


1
AEC CR i OC i
 
( ) ( )
CR(i)
OC(i)
AEC
n*

14. An asset purchased 3 years ago is now challenged by a new piece of equipment.
The present market value of the defender is Rs.130000. anticipated salvage values
and Annual Operating Costs (AOC) for the next 5 years are given in the table.
What is the minimum cost life to be used while comparing this defender with a
challenger if a 10% year return is required.
Life in years Salvage value AOC
1 Rs 90,000 Rs 25,000
2 Rs 80,000 Rs 27,000
3 Rs 60,000 Rs 30,000
4 Rs 20,000 Rs 35,000
5 Rs 0.00 Rs 45,000

15. CR(i) = (P-F) (A/P, i, n) + Fi
Finding for n=1, 2, 3, 4, 5
n=1,
CR(i) = (1,30,000-90,000) (A/P, 10,1) + 90000 x 0.1= 53000
n=2,
CR= (1,30,000-80,000) (A/P, 10,2) + 80000 x 0.1 = 36810
n=3,
CR= (1,30,000-60,000) (A/P, 10,3) + 60000 x 0.1 = 34147

16. CR(i) = (P-F) (A/P, i, n) + Fi
n=4,
CR= (1,30,000-20,000) (A/P, 10,4) + 20000 x 0.1 = 36705
n=5,
CR= (1,30,000-0) (A/P, 10,5) + 0 = 34294

17. Equivalent Annual Operating Costs for n= 1,2,3,4,5
n= 1, A= 25,000
n=2,
A= [25000 (P/F,10,1) + 27000 (P/F,10,2)] x (A/P, 10,2)
= 25952
n=3,
A= [25000 (P/F,10,1) + 27000 (P/F,10,2) + 30000 (P/F, 10,3)] x (A/P, 10,3)
= 27174

18. Equivalent Annual Operating Costs for n= 1,2,3,4,5
n=4,
[25000 (P/F,10,1) +27000 (P/F,10,2) +30000 (P/F,10,3) +35000 (P/F, 10,4)] x (A/P, 10,4)
= 28861
n=5
A= [25000 (P/F,10,1) + 27000(P/F,10,2) + 30000 (P/F,10,3) + 35000 (P/F, 10,4) +
45000 (P/F,10,5)] x (A/P, 10,5)
= 31504

19. Year CR (i) AOC EUAC
1 53000 25000 78000
2 36810 25952 62322
3 34148 27174 61322
4 36702 28861 65563
5 34294 31504 65798
Minimum total EUAC occur at year 3.
Hence economic life of the asset is 3 years